Solution
= Solution
Along the first <sequence>, $\sin(1/x_n)=\sin(2n\pi+\pi/6)=1/2$, whereas along the second <sequence>, $\sin(1/x'_n)=\sin(2n\pi)=0$. The outer <sine> therefore gives
$$
\boxed{f_0(x_n)=\sin(\pi/2)=1,\qquad f_0(x'_n)=0}.
$$
Both <sequences> approach zero through positive values. For arbitrary $a$, the same calculation gives $f_a(x_n)=x_n^a$ and $f_a(x'_n)=0$.